c上的二进制搜索，while循环

在C语言中，我无法从二进制搜索代码中得到一些东西

int binarySearch(int a[], int n, int x)
{
   int low=0, mid, high=n-1;
   while(low <= high)
   {
        mid = (low + high) / 2;
        if (x < a[mid])
            high = mid - 1;
        else if (x > a[mid])
         low = mid + 1;
       else
         return mid;
   }
   return -1;
}

intbinarysearch（inta[]，intn，intx）
{
int低=0，中，高=n-1；
while（低a[中]）
低=中+1；
其他的
中途返回；
}
返回-1；
}

为什么while循环

while（left采用一个简单的

int a[1] = {5};
printf("%d\n", binarySearch(a, 1, 5));

使用
while（低<高）
，代码打印-1（未找到-回答错误）

使用
while（低部分1-快速回答此特定问题
这不是对二进制搜索的完整总结，但我将简而言之以解决这个问题。

这两个while循环条件的关键区别，给定

1.相应地更新
低位
（
左侧
）和
高位
（
右侧
）指针。通过“相应地”，请参见第3部分。

2.假设
低
（
左
）和
高
（
右
）

3.假设目标存在于数组中


while（low）如果没有完整的代码，这很难说。Iv'e添加了代码。我看到了，我看到它不起作用，但我想知道为什么，谢谢。你真的在问“是
@Olaf吗？好吧，我能想到两种变体的算法表现相同……所以问题是”算法是否与
相同？使用一个元素数组在代码中单步执行肯定会说明为什么

    public int binarySearch(int[] nums, int target){

        int left = 0, right = nums.length - 1;
        // please pay attention to the initial condition of right(ptr)
        while(left <= right){
            // to floor the mid
            int mid = left + (right - left) / 2;

            // to check whether the middle element is equal to the target in every iteration
            if(nums[mid] == target) return mid;
            else if(target > nums[mid]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return -1;
    }

    public int binarySearch(int[] nums, int target){
        // please pay attention to the initial condition or the right(ptr)
        int left = 0, right = nums.length;

        while(left < right){
            int mid = left + (right - left) / 2;

            // please pay twice attention to the equality case
            if(target > nums[mid]) {
                left = mid + 1;
             } else {
                right = mid;
             }
        }

        return left;
    }

    public int binarySearchWithFlooringMid(int[] nums, int target){
        // please pay attention to the initial condition of right(ptr)
        int left = 0, right = nums.length - 1;
        while(left <= right){
            // to floor the mid
            int mid = left + (right - left) / 2;

            // to check whether the middle element is equal to the target in every iteration
            if(nums[mid] == target) return mid;
            else if(target > nums[mid]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return -1;
    }

    public int binarySearchWithCeilingMid(int[] nums, int target){
        int left = 0, right = nums.length - 1;
        while(left <= right){
            // to ceil the mid
            int mid = left + (right - left + 1) / 2;

            // to check whether the middle element is equal to the target in every iteration
            if(nums[mid] == target) return mid;
            else if(target > nums[mid]) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }

        return -1;
    }

    public int binarySearchLeftmost(int[] nums, int target){
        int left = 0, right = nums.length;

        while(left < right){
            int mid = left + (right - left) / 2;

            // please pay twice attention to the equality case
            if(target > nums[mid]) {
                left = mid + 1;
             } else {
                 right = mid;
             }
        }

        return left;
    }

    public int binarySearchRightmost(int[] nums, int target){
        int left = 0, right = nums.length;

        while(left < right){
            int mid = left + (right - left) / 2;

            // please pay twice attention to the equality case
            if(target < nums[mid]) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        return right - 1;
    }